From a circular disc of radius R, a square is cut out with a radius as its diagonal. The center of mass of remaining portion is at a distance from the center)

Options

A. `R/(4pi-2)`
B. `R/(2pi)`
C. `R/(pi-2)`
D. `R/(2pi -2)`

Explanation

To solve the problem of finding the distance of the center of mass of the remaining portion of a circular disc after cutting out a square, we can follow these steps:

### Step 1: Understand the Geometry
We have a circular disc of radius \( R \) and a square is cut out such that the diagonal of the square is equal to the radius \( R \) of the disc.

### Step 2: Determine the Side Length of the Square
The diagonal \( d \) of the square is given by:
\[ d = R \]
Using the relationship between the side length \( a \) of the square and its diagonal:
\[ d = a\sqrt{2} \implies a = \frac{R}{\sqrt{2}} \]

### Step 3: Calculate the Area of the Square
The area \( A_s \) of the square can be calculated as:
\[ A_s = a^2 = \left(\frac{R}{\sqrt{2}}\right)^2 = \frac{R^2}{2} \]

### Step 4: Calculate the Mass of the Disc and the Square
Assuming the density of the material is \( \rho \):
- The mass \( M_d \) of the disc is:
\[ M_d = \rho \cdot \text{Area of the disc} = \rho \cdot \pi R^2 \]
- The mass \( M_s \) of the square is:
\[ M_s = \rho \cdot A_s = \rho \cdot \frac{R^2}{2} \]

### Step 5: Set Up the Coordinate System
Assume the center of the disc is at the origin (0,0). The center of the square, which is cut out, will be at:
\[ \left(\frac{R}{2\sqrt{2}}, 0\right) \]

### Step 6: Calculate the Center of Mass of the Remaining Portion
Using the formula for the center of mass of a system of particles:
\[ x_{cm} = \frac{M_d \cdot x_d - M_s \cdot x_s}{M_d - M_s} \]
Substituting the values:
- \( x_d = 0 \) (center of the disc)
- \( x_s = \frac{R}{2\sqrt{2}} \)

We have:
\[ x_{cm} = \frac{(\rho \pi R^2) \cdot 0 - \left(\rho \frac{R^2}{2}\right) \cdot \left(\frac{R}{2\sqrt{2}}\right)}{M_d - M_s} \]
\[ = \frac{-\left(\frac{\rho R^3}{4\sqrt{2}}\right)}{\rho \pi R^2 - \frac{\rho R^2}{2}} \]

### Step 7: Simplify the Expression
The denominator simplifies to:
\[ \rho R^2 \left(\pi - \frac{1}{2}\right) \]
Thus, we have:
\[ x_{cm} = \frac{-\frac{R^3}{4\sqrt{2}}}{R^2 \left(\pi - \frac{1}{2}\right)} = \frac{-R}{4\sqrt{2} \left(\pi - \frac{1}{2}\right)} \]

### Step 8: Final Result
The distance of the center of mass of the remaining portion from the center of the disc is:
\[ \text{Distance} = \frac{R}{4\sqrt{2} \left(\pi - \frac{1}{2}\right)} \]